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‎README.adoc‎

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@@ -5932,14 +5932,14 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
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//|{doc_base_url}/0843-guess-the-word.adoc[题解]
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//|Hard
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//|
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//
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//|{counter:codes}
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//|{leetcode_base_url}/backspace-string-compare/[844. Backspace String Compare^]
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//|{source_base_url}/_0844_BackspaceStringCompare.java[Java]
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//|{doc_base_url}/0844-backspace-string-compare.adoc[题解]
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//|Easy
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//|
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//
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|{counter:codes}
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|{leetcode_base_url}/backspace-string-compare/[844. Backspace String Compare^]
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|{source_base_url}/_0844_BackspaceStringCompare.java[Java]
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|{doc_base_url}/0844-backspace-string-compare.adoc[题解]
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|Easy
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|
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//|{counter:codes}
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//|{leetcode_base_url}/longest-mountain-in-array/[845. Longest Mountain in Array^]
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//|{source_base_url}/_0845_LongestMountainInArray.java[Java]

‎docs/0844-backspace-string-compare.adoc‎

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[#0844-backspace-string-compare]
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= 844. Backspace String Compare
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= 844. 比较含退格的字符串
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{leetcode}/problems/backspace-string-compare/[LeetCode - Backspace String Compare^]
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https://leetcode.cn/problems/backspace-string-compare/[LeetCode - 844. 比较含退格的字符串^]
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Given two strings `S` and `T`, return if they are equal when both are typed into empty text editors. `#` means a backspace character.
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给定 `s` 和 `t` 两个字符串，当它们分别被输入到空白的文本编辑器后，如果两者相等，返回 `true`。`#` 代表退格字符。
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**注意：**如果对空文本输入退格字符，文本继续为空。
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*Example 1:*
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*示例 1：*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* S = "ab#c", T = "ad#c"
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*Output:* true
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*Explanation*: Both S and T become "ac".
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----
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*Example 2:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* S = "ab##", T = "c#d#"
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*Output:* true
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*Explanation*: Both S and T become "".
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----
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*Example 3:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* S = "a##c", T = "#a#c"
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*Output:* true
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*Explanation*: Both S and T become "c".
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----
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*Example 4:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* S = "a#c", T = "b"
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*Output:* false
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*Explanation*: S becomes "c" while T becomes "b".
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----
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....
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输入：s = "ab#c", t = "ad#c"
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输出：true
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解释：s 和 t 都会变成 "ac"。
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....
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*Note*:
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*示例 2：*
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....
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输入：s = "ab##", t = "c#d#"
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输出：true
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解释：s 和 t 都会变成 ""。
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....
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* `1 <= S.length <= 200`
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* `1 <= T.length <= 200`
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* `S` and `T` only contain lowercase letters and `'#'` characters.
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*示例 3：*
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....
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输入：s = "a#c", t = "b"
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输出：false
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解释：s 会变成 "c"，但 t 仍然是 "b"。
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....
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*Follow up:*
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* Can you solve it in `O(N)` time and `O(1)` space?
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*提示：*
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* `1 \<= s.length, t.length \<= 200`
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* `s` 和 `t` 只含有小写字母以及字符 `#`
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*进阶：*
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* 你可以用 stem:[O(n)] 的时间复杂度和 stem:[O(1)] 的空间复杂度解决该问题吗？
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== 思路分析
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如果不想占用额外的空间，可以从后向前比较：如果双方都是字母，则直接比较；否则将不是字母的下标向前推进到字母或者结束为止。
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image::images/0844-10.gif[{image_attr}]
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[[src-0844]]
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[tabs]
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====
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一刷::
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0844_BackspaceStringCompare.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0844_BackspaceStringCompare_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/backspace-string-compare/solutions/451606/bi-jiao-han-tui-ge-de-zi-fu-chuan-by-leetcode-solu/[844. 比较含退格的字符串 - 官方题解^]
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. https://leetcode.cn/problems/backspace-string-compare/solutions/683776/shuang-zhi-zhen-bi-jiao-han-tui-ge-de-zi-8fn8/[844. 比较含退格的字符串 - 【双指针】比较含退格的字符串^]
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. https://leetcode.cn/problems/backspace-string-compare/solutions/451967/844zhan-mo-ni-yu-kong-jian-geng-you-de-shuang-zhi-/[844. 比较含退格的字符串 - 【栈模拟】与【空间更优的双指针】详解^]

‎docs/images/0844-10.gif‎

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‎docs/index.adoc‎

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// include::0844-backspace-string-compare.adoc[leveloffset=+1]
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include::0844-backspace-string-compare.adoc[leveloffset=+1]
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// include::0845-longest-mountain-in-array.adoc[leveloffset=+1]
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‎logbook/202601.adoc‎

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|✅ 模拟计算题。
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|{counter:codes}
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|{leetcode_base_url}/split-array-into-fibonacci-sequence/[842. Split Array into Fibonacci Sequence^]
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|{leetcode_base_url}/split-array-into-fibonacci-sequence/[842. 将数组拆分成斐波那契序列^]
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|{doc_base_url}/0842-split-array-into-fibonacci-sequence.adoc[题解]
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|✅ 回溯。将字符串分割成数字，逐个检查是否可以组成斐波那契数列，不符合要求就回溯。
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|{counter:codes}
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|{leetcode_base_url}/backspace-string-compare/[844. 比较含退格的字符串^]
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|{doc_base_url}/0844-backspace-string-compare.adoc[题解]
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|✅ 双指针。从后向前比较：如果双方都是字母，则直接比较；否则将不是字母的下标向前推进到字母或者结束为止。
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|===
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‎src/main/java/com/diguage/algo/leetcode/_0844_BackspaceStringCompare.java‎

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package com.diguage.algo.leetcode;
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public class _0844_BackspaceStringCompare {
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// tag::answer[]
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/**
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* 代码写的好丑！
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*
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* @author D瓜哥 · https://www.diguage.com
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* @since 2026-08-01 21:10:29
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*/
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public boolean backspaceCompare(String s, String t) {
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int si = s.length() - 1;
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int ti = t.length() - 1;
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int sj = 0, tj = 0;
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while (si >= 0 || ti >= 0) {
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if (si >= 0 && s.charAt(si) != '#'
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&& ti >= 0 && t.charAt(ti) != '#'
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&& s.charAt(si) != t.charAt(ti)) {
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return false;
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}
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while (sj > 0 || (si >= 0 && s.charAt(si) == '#')) {
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if (si >= 0 && s.charAt(si) != '#' && sj > 0) {
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si--;
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sj--;
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} else if (si >= 0 && s.charAt(si) == '#') {
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si--;
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sj++;
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}
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if (si < 0) {
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break;
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}
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}
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while (tj > 0 || (ti >= 0 && t.charAt(ti) == '#')) {
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if (ti >= 0 && t.charAt(ti) != '#' && tj > 0) {
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ti--;
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tj--;
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} else if (ti >= 0 && t.charAt(ti) == '#') {
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ti--;
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tj++;
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}
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if (ti < 0) {
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break;
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}
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}
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if (si >= 0 && ti >= 0 && s.charAt(si) != t.charAt(ti)) {
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return false;
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} else if (si < 0 && ti < 0) {
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break;
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} else if (si < 0 || ti < 0) {
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return false;
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}
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si--;
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ti--;
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}
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return true;
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}
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// end::answer[]
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static void main() {
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new _0844_BackspaceStringCompare()
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.backspaceCompare("nzp#o#g", "b#nzp#o#g");
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// .backspaceCompare("bxj##tw", "bxj###tw");
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// .backspaceCompare("a##c", "#a#c");
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// .backspaceCompare("ab##", "c#d#");
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// .backspaceCompare("ab#c", "ad#c");
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}
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}

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